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How much heat (kJ) is absorbed by 825.4 g of water in order for the temperature to increase from 25.00°C to 32.50°C?

Respuesta :

Eduard22sly

The heat absorbed by the water is 25.9 KJ

To solve the question given above, we'll begin by calculating the change in the temperature of the water. This can be obtained as follow:

Initial temperature (T₁) = 25 °C  

Final temperature (T₂) = 32.50 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 32.50 – 25

ΔT = 7.5 °C

Finally, we shall determine the heat absorbed by the water. This can be obtained as follow:

Mass of water = 825.4 g

Change in temperature (ΔT) = 7.5 °C

Specific heat capacity of water = 4.184 J/gºC

Heat absorbed (Q) =?

Q = MCΔT

Q = 825.4 × 4.184 × 7.5

Q = 25901.052 J

Divide by 1000 to express in KJ

Q = 25901.052 / 1000

Q = 25.9 KJ

Therefore, the heat absorbed by the water is 25.9 KJ.

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